3 years ago

A 5’5 person weight 4.4kg. How many pounds is this

Mathematics

1 answer:

marshall27 [118]3 years ago

6 0

Answer:

9.70034

Step-by-step explanation:

1kg=2.20462 pounds, just do the math! I hope this helped.

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Solve for x: 3-(2x-5) < -4(x+2)

Hitman42 [59]

Answer:

r>5

Step-by-step explanation:

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3 years ago

Please help me with this...

mr_godi [17]

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A. A car travels 20 m/s and reached its destination in a record time of 10 secs. Calculate the distance

Doss [256]

The car travelled 200 meters

20 m/s times 10 = 200 m in 10 s

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2 years ago

Please help! i need to pass this class!!

torisob [31]

Angle B = 54°

Step-by-step explanation:

complimentary means they add up to 90

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also x= 51

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3 years ago

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Consider the differential equation y'' − y' − 30y = 0. Verify that the functions e−5x and e6x form a fundamental set of solution

Alex Ar [27]

Answer:

Step-by-step explanation:

We have to take the derivatives for both functions and replace in the differential equation. Hence

for y=e^{-5x}:

$y(x)=e^{-5x}\\y'(x)=-5e^{-5x}\\y''(x)=25e^{-5x}\\$

for y=e^{6x}:

$y(x)=e^{6x}\\y'(x)=6e^{6x}\\y''(x)=36e^{6x}\\$

Now we replace in the differential equation y'' − y' − 30y = 0

for y=e^{-5x}:

$25e^{-5x}+5e^{-5x}-30e^{-5x}=0\\25+5-30=0$

for y=e^{6x}:

$36e^{6x}-6e^{6x}-30=0\\36-6+30=0$

Now, to know if both function are linearly independent we calculate the Wronskian

$W(f,g)=fg'-f'g$

$W(e^{-5x},e^{6x})=(e^{-5x})(6e^{6x})-(-5e^{-5x})(e^{6x})\neq 0$

I hope this is useful for you

Best regard

7 0

3 years ago